数学建模第三次作业追击问题Word文档格式.doc
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数学建模第三次作业追击问题Word文档格式.doc
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(1)四个人能否追到一起?
(2)若能追到一起,则每个人跑过多少路程?
(3)追到一起所需要的时间(设速率为1)?
(4)如果四个人追逐的速度不一样,情况又如何呢
分析:
先建立坐标系,设计程序使从A,B,C,D四个点同时出发,画出图形并判断。
程序设计流程:
四个人追击的速度相等,则有。
针对这种情形,可有以下的程序。
holdon
axis([0202]);
grid
A=[0,0];
B=[0,1];
C=[1,1];
D=[1,0];
k=0;
s1=0;
s2=0;
s3=0;
s4=0;
%四个人分别走过的路程
t=0;
v=1;
dt=0.002;
whilek<
10000
k=k+1;
plot(A
(1),A
(2),'
r.'
'
markersize'
15);
plot(B
(1),B
(2),'
b.'
plot(C
(1),C
(2),'
m.'
plot(D
(1),D
(2),'
k.'
e1=B-A;
d1=norm(e1);
e2=C-B;
d2=norm(e2);
e3=D-C;
d3=norm(e3);
e4=A-D;
d4=norm(e4);
fprintf('
k=%.0f'
k)
A(%.2f,%.2f)d1=%.2f'
A
(1),A
(2),d1)
B(%.2f,%.2f)d2=%.2f'
B
(1),B
(2),d2)
C(%.2f,%.2f)d3=%.2f'
C
(1),C
(2),d3)
D(%.2f,%.2f)d4=%.2f\n'
D
(1),D
(2),d4)
A=A+v*dt*e1/d1;
B=B+v*dt*e2/d2;
C=C+v*dt*e3/d3;
D=D+v*dt*e4/d4;
t=t+dt;
s1=s1+v*dt;
s2=s2+v*dt;
s3=s3+v*dt;
s4=s4+v*dt;
ifnorm(A-C)<
=5.0e-3&
norm(B-D)<
=5.0e-3
break
end
end
t
s1
s2
s3
s4
部分运行结果:
k=481A(0.52,0.52)d1=0.04B(0.52,0.48)d2=0.04C(0.48,0.48)d3=0.04D(0.48,0.52)d4=0.04
k=482A(0.52,0.52)d1=0.04B(0.52,0.48)d2=0.04C(0.48,0.48)d3=0.04D(0.48,0.52)d4=0.04
k=483A(0.52,0.52)d1=0.04B(0.52,0.48)d2=0.04C(0.48,0.48)d3=0.04D(0.48,0.52)d4=0.04
k=484A(0.52,0.51)d1=0.04B(0.51,0.48)d2=0.04C(0.48,0.49)d3=0.04D(0.49,0.52)d4=0.04
k=485A(0.52,0.51)d1=0.04B(0.51,0.48)d2=0.04C(0.48,0.49)d3=0.04D(0.49,0.52)d4=0.04
k=486A(0.52,0.51)d1=0.03B(0.51,0.48)d2=0.03C(0.48,0.49)d3=0.03D(0.49,0.52)d4=0.03
k=487A(0.52,0.51)d1=0.03B(0.51,0.48)d2=0.03C(0.48,0.49)d3=0.03D(0.49,0.52)d4=0.03
k=488A(0.52,0.51)d1=0.03B(0.51,0.48)d2=0.03C(0.48,0.49)d3=0.03D(0.49,0.52)d4=0.03
k=489A(0.52,0.51)d1=0.03B(0.51,0.48)d2=0.03C(0.48,0.49)d3=0.03D(0.49,0.52)d4=0.03
k=490A(0.52,0.50)d1=0.03B(0.50,0.48)d2=0.03C(0.48,0.50)d3=0.03D(0.50,0.52)d4=0.03
k=491A(0.52,0.50)d1=0.02B(0.50,0.48)d2=0.02C(0.48,0.50)d3=0.02D(0.50,0.52)d4=0.02
k=492A(0.52,0.50)d1=0.02B(0.50,0.48)d2=0.02C(0.48,0.50)d3=0.02D(0.50,0.52)d4=0.02
k=493A(0.51,0.50)d1=0.02B(0.50,0.49)d2=0.02C(0.49,0.50)d3=0.02D(0.50,0.51)d4=0.02
k=494A(0.51,0.50)d1=0.02B(0.50,0.49)d2=0.02C(0.49,0.50)d3=0.02D(0.50,0.51)d4=0.02
k=495A(0.51,0.50)d1=0.02B(0.50,0.49)d2=0.02C(0.49,0.50)d3=0.02D(0.50,0.51)d4=0.02
k=496A(0.51,0.50)d1=0.01B(0.50,0.49)d2=0.01C(0.49,0.50)d3=0.01D(0.50,0.51)d4=0.01
k=497A(0.51,0.49)d1=0.01B(0.49,0.49)d2=0.01C(0.49,0.51)d3=0.01D(0.51,0.51)d4=0.01
k=498A(0.51,0.49)d1=0.01B(0.49,0.49)d2=0.01C(0.49,0.51)d3=0.01D(0.51,0.51)d4=0.01
k=499A(0.50,0.49)d1=0.01B(0.49,0.50)d2=0.01C(0.50,0.51)d3=0.01D(0.51,0.50)d4=0.01
k=500A(0.50,0.50)d1=0.01B(0.50,0.50)d2=0.01C(0.50,0.50)d3=0.01D(0.50,0.50)d4=0.01
k=501A(0.50,0.50)d1=0.01B(0.50,0.50)d2=0.01C(0.50,0.50)d3=0.01D(0.50,0.50)d4=0.01
k=502A(0.50,0.50)d1=0.00B(0.50,0.50)d2=0.00C(0.50,0.50)d3=0.00D(0.50,0.50)d4=0.00
t=
1.0040
s1=
s2=
s3=
s4=
1.0040
从运行的结果来看,如果四个人的追击速度相同,均为1,可有以下的结果:
(1)四人最后可以追到一起。
(2)每个人跑过相等的路程,均为1.0040.
(3)追到一起的时间为1.0040秒。
如果四个人追击的速度不一样,可取,运行程序。
运行程序:
dt=0.001;
B=B+2*v*dt*e2/d2;
C=C+2*v*dt*e3/d3;
=1.0e-3&
=1.0e-3
k=662A(0.41,0.32)d1=0.06B(0.43,0.27)d2=0.00C(0.43,0.27)d3=0.00D(0.43,0.27)d4=0.05
k=663A(0.41,0.32)d1=0.05B(0.43,0.27)d2=0.00C(0.43,0.27)d3=0.00D(0.43,0.27)d4=0.05
k=664A(0.41,0.32)d1=0.05B(0.43,0.27)d2=0.00C(0.43,0.27)d3=0.00D(0.43,0.27)d4=0.05
k=665A(0.41,0.32)d1=0.05B(0.43,0.27)d2=0.00C(0.43,0.27)d3=0.00D(0.43,0.28)d4=0.05
k=666A(0.41,0.31)d1=0.05B(0.43,0.28)d2=0.00C(0.43,0.28)d3=0.00D(0.43,0.28)d4=0.04
k=667A(0.41,0.31)d1=0.04B(0.43,0.28)d2=0.00C(
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- 数学 建模 第三次 作业 追击 问题